Let X_{1},\cdots,X_{n} be a random sample from f(x∣\theta)=\thetx^{\theta-1} for 0<x<1.

dooporpplauttssg 2022-04-29 Answered

Let X1,,Xn be a random sample from f(xθ)=θxθ1 for 0<x<1. Find a confidence interval for θ using as pivotal quantity a function of the maximum likelihood estimator for θ.

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Answers (1)

Makayla Santiago
Answered 2022-04-30 Author has 19 answers
Step 1
Assuming you want a confidence interval for θ, you may use the CLT or a known chi distribution for variances and some integrals to get an interval for the parameter. First, we have that
θ^=nln xi.
For the expected value we have the integral
E[X]=01θx(θ1)x dθ=θ1+θ
And for the second moment
μ2=01θxθ1x2 dθ=θ2+θ
So the variance is
Now then, we know that the following quotient has a chi-squared distribution, so in this case
C=(n1)S2σ2χ{n1}2  and  C=(n1)S2θ(θ+1)2(θ+2)
where S2=1n1(XiX)2. Finally we can work with the random variable C distributed as a chi squared. We want a confidence interval of (1α). Then
denoting that χ{n1}2(β) is the β percentile of the χ2 which may be found in reference tables.
From here we work with the interval only substituting the percentiles with 1Dt and 1Dπ, lower and superior.
where, taking the before definition and derivation, θ^=nln xi and S2=1n1(XiX)2. When we sent the θ to the other side, we simply used its estimate to work around the problem. (However a mathematician should confirm this is valid!) Either way, if you find a better known distribution that may include θ, then try it and tell us if it is simpler!
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